STOICHIOMETRY
TUTORIAL
Paul Gilletti
1
Instructions: This is a work along tutorial. Each time you click
the mouse or touch the space bar on your computer, one step of
the problem solving occurs. Pressing the PAGE UP key will
backup the steps.
Get a pencil and
paper, a periodic
table and a
calculator, and
let’s get to work.
2
(1-2-3) General Approach For Problem Solving:
1. Clearly identify the Goal or Goals and the UNITS involved. (What
are you trying to do?)
2. Determine what is given and the UNITS.
3. Use conversion factors (which are really ratios) and their UNITS to
CONVERT what is given into what is desired.
3
Table of Contents: Click on each tab to view problem types.
View Complete Slide Show
Sample problem 1
Sample problem 2
Converting grams to moles
Mole to Mole Conversions
Gram-Mole and Gram-Gram Problems
Solution Stoichiometry Problems
Limiting/Excess/ Reactant and Theoretical Yield Problems :
4
Sample problem for general problem solving.
Sam has entered into a 10 mile marathon. Use ALL of the following
conversions (ratios) to determine how many inches there are in the race.
5280 ft = 1 mile; 12 inches = 1 ft
1. What is the goal and what units are needed?
Goal = ______ inches
2. What is given and its units?
10 miles
3. Convert using factors (ratios).
10 miles
Given
5280 ft
1 mile
12 inches
1 ft
Units match
=
633600 inches
Goal
Convert
5
Menu
Sample problem #2 on problem solving.
A car is traveling at a speed of 45 miles per hr (45 miles/hr). Determine
its speed in kilometers per second using the following conversion
factors (ratios). 1 mile = 5280 ft; 1 ft = 12 in; 1 inch = 2.54 cm; k = 1 x
103; c = 1 x 10-2; 1 hr =60 min; 1 min = 60 s
Goal
Given
-2
k
hr min
45 mi 5280 ft 12 in 2.54 cm 1 x 10
mi
hr
c
1 ft
1 x 10 3 60 min 60 s
1 in
c cancels c
m remains
This is
the
same as
putting
k over k
= 0.020
km
s
Units Match!
6
Converting grams to moles.
Determine how many moles there are in 5.17 grams of Fe(C5H5)2.
Given
5.17 g Fe(C5H5)2
Goal
units match
mol
185.97 g
Use the molar mass to
convert grams to
moles.
= 0.0278
moles Fe(C5H5)2
Fe(C5H5)2
2 x 5 x 1.001 = 10.01
2 x 5 x 12.011 = 120.11
1 x 55.85 = 55.85
185.97 g
mol
7
Stoichiometry (more working with ratios)
Ratios are found within a chemical equation.
2HCl + 1Ba(OH)2  2H2O + 1 BaCl2
coefficients give MOLAR RATIOS
2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O
and 1 mole of BaCl2
8
Mole – Mole Conversions
When N2O5 is heated, it decomposes:
2N2O5(g)  4NO2(g) + O2(g)
a. How many moles of NO2 can be produced from 4.3 moles of N2O5?
2N2O5(g)  4NO2(g) + O2(g)
4.3 mol
? mol Units match
4.3 mol N2O5
4mol NO 2
2mol N 2O 5
= 8.6
moles NO2
b. How many moles of O2 can be produced from 4.3 moles of N2O5?
2N2O5(g)  4NO2(g) + O2(g)
4.3 mol
4.3 mol N2O5
1mol O 2
2mol N 2O 5
? mol
= 2.2
mole O2
9
gram ↔ mole and gram ↔ gram conversions
When N2O5 is heated, it decomposes:
2N2O5(g)  4NO2(g) + O2(g)
a. How many moles of N2O5 were used if 210g of NO2 were produced?
2N2O5(g)  4NO2(g) + O2(g)
? moles
210g
210 g NO2
mol NO 2
46.0g NO 2
Units match
2mol N 2O 5
4mol NO 2
= 2.28
moles N2O5
b. How many grams of N2O5 are needed to produce 75.0 grams of O2?
2N2O5(g)  4NO2(g) + O2(g)
75.0 g
? grams
75.0 g O2
mol O 2
32.0 g O 2
2mol N 2O 5
1mol O 2
108g N 2O 5
mol N 2O 5
= 506 grams N2O5
10
Gram to Gram Conversions
Aluminum is an active metal that when placed in hydrochloric acid produces
hydrogen gas and aluminum chloride. How many grams of aluminum
chloride can be produced when 3.45 grams of aluminum are reacted with an
excess of hydrochloric acid?
2 Al(s) + 6HCl(aq)  2AlCl3(aq) +
3 H2(g)
First write a balanced
equation.
11
Gram to Gram Conversions
Aluminum is an active metal that when placed in hydrochloric acid produces
hydrogen gas and aluminum chloride. How many grams of aluminum
chloride can be produced when 3.45 grams of aluminum are reacted with an
excess of hydrochloric acid?
2 Al(s) + 6HCl(aq)  2AlCl3(aq) +
? grams
3.45 g
3 H2(g)
Now let’s get organized.
Write the information
below the substances.
12
gram to gram conversions
Aluminum is an active metal that when placed in hydrochloric acid produces
hydrogen gas and aluminum chloride. How many grams of aluminum
chloride can be produced when 3.45 grams of aluminum are reacted with an
excess of hydrochloric acid?
2 Al(s) + 6HCl(aq)  2AlCl3(aq) +
? grams
3.45 g
3 H2(g)
Units match
3.45 g Al
mol Al
27.0 g Al
2 mol AlCl 3 133.3 g AlCl 3
mol AlCl 3
2 mol Al
Now
We must
Now
use
Let’s
the
always
use
work
molar
thethe
convert
molar
mass
problem.
ratio.
to
toconvert
moles.
to grams.
=
17.0 g AlCl3
13
14
Molarity
Molarity is a term used to express concentration. The units of molarity are
moles per liter (It is abbreviated as a capital M)
When working problems, it
is a good idea to change M
into its units.
moles
moles
M

Liter 1000 mL
15
16
Solutions
A solution is prepared by dissolving 3.73 grams of AlCl3 in
water to form 200.0 mL solution. A 10.0 mL portion of the
solution is then used to prepare 100.0 mL of solution.
Determine the molarity of the final solution.
What type of
problem(s) is
this?
Molarity
followed by
dilution.
17
Solutions
A solution is prepared by dissolving 3.73 grams of AlCl3 in
water to form 200.0 mL solution. A 10.0 mL portion of the
solution is then used to prepare 100.0 mL of solution.
Determine the molarity of the final solution.
1st:
3.73 g
mol
= 0.140 mol
3
133.4 g 200.0 x 10 L
L
molar mass of AlCl3
2nd:
M1V1 = M2V2
dilution formula
(0.140 M)(10.0 mL) = (? M)(100.0 mL)
0.0140 M = M2 final concentration
18
19
Solution Stoichiometry
50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and
solid NaHCO3 (baking soda) is to be used to neutralize the
acid. How many grams of NaHCO3 must be used?
H2SO4(aq) + 2NaHCO3  2H2O(l) + Na2SO4(aq) + 2CO2(g)
20
Solution Stoichiometry
50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and
solid NaHCO3 (baking soda) is to be used to neutralize the
acid. How many grams of NaHCO3 must be used?
H2SO4(aq) + 2NaHCO3  2H2O(l) + Na2SO4(aq) + 2CO2(g)
50.0 mL
? g Our Goal
6.0 M
=
6.0 mol
L
Look!
A conversion
factor!
21
Solution Stoichiometry
50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and
solid NaHCO3 (baking soda) is to be used to neutralize the
acid. How many grams of NaHCO3 must be used?
H2SO4(aq) + 2NaHCO3  2H2O(l) + Na2SO4(aq) + 2CO2(g)
50.0 mL
? g Our Goal
6.0 M
=
6.0 mol
L
H2SO4
50.0 mL 6.0 mol H 2SO 4
1000mL
H 2SO 4
NaHCO3 NaHCO3
84.0 g
2 mol
= 50.4 g NaHCO3
mol
1 mol
NaHCO3
H2SO4
22
23
Solution Stoichiometry:
Determine how many mL of 0.102 M NaOH solution are needed to
neutralize 35.0 mL of 0.125 M H2SO4 solution.
2
1 2SO4 
____NaOH
+ ____H
2 2O
____H
1 2SO4
+ ____Na
First write a balanced
Equation.
24
Solution Stoichiometry:
Determine how many mL of 0.102 M NaOH solution is needed to neutralize
35.0 mL of 0.125 M H2SO4 solution.
2
1 2SO4 
____NaOH
+ ____H
0.102 M mol
L
? mL
Our Goal
2 2O
____H
1 2SO4
+ ____Na
35.0 mL
0.125 mol 0.125 mol

L
1000 mL
Since 1 L = 1000 mL, we can use
this to save on the number of conversions
Now, let’s get organized. Place
numerical Information and
accompanying UNITS below each
compound.
25
Solution Stoichiometry:
Determine how many mL of 0.102 M NaOH solution is needed to neutralize
35.0 mL of 0.125 M H2SO4 solution.
2
1 2SO4 
____NaOH
+ ____H
0.102 M mol
L
? mL
H2SO4
35.0 mL
2 2O
____H
1 2SO4
+ ____Na
35.0 mL
0.125 mol 0.125 mol

L
1000mL
H2SO4
0.125 mol
1000 mL
H2SO4
NaOH
2 mol
1 mol
H2SO4
Units Match
1000 mL NaOH = 85.8 mL NaOH
0.102 mol NaOH
Now let’s get to work
converting.
26
27
Solution Stoichiometry
What volume of 0.40 M HCl solution is needed to
completely neutralize 47.1 mL of 0.75 M Ba(OH)2?
1st write out
a balanced chemical
equation
28
Solution Stoichiometry
What volume of 0.40 M HCl solution is needed to
completely neutralize 47.1 mL of 0.75 M Ba(OH)2?
2HCl(aq) +
Ba(OH)2(aq) 
0.40 M
47.1 mL
0.75 M
? mL
Ba(OH)2
47.1 mL
0.75mol Ba(OH)2
1000 mL Ba(OH)2
2H2O(l) + BaCl2
Units match
HCl
2 mol
1 mol
Ba(OH)2
HCl
1000 mL
0.40 mol
HCl
= 176 mL HCl
29
30
31
Solution Stochiometry Problem:
A chemist performed a titration to standardize a barium hydroxide solution.
If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of
the barium hydroxide solution, what was the concentration of the barium
hydroxide solution in moles per liter (M)?
2
1
2 2O(l) + ____BaCl
1
____HCl(aq)
+ ____Ba(OH)
2(aq)  ____H
2(aq)
25.00 mL
23.28 mL
0.135 mol
L
? mol
L
First write a balanced
chemical reaction.
32
Solution Stochiometry Problem:
A chemist performed a titration to standardize a barium hydroxide solution.
If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of
the barium hydroxide solution, what was the concentration of the barium
hydroxide solution in moles per liter (M)?
2
1
2 2O(l) + ____BaCl
1
____HCl(aq)
+ ____Ba(OH)
2(aq)  ____H
2(aq)
25.00 mL
23.28 mL
Units match on top!
? mol
0.135 mol
L
L
23.28 mL HCl
25.00 x 10-3 L
Ba(OH)2
0.135 mol HCl l mol Ba(OH)2
= 0.0629 mol Ba(OH)2
1000 mL HCl 2 mol HCl
L Ba(OH)
2
Units Already Match on Bottom!
33
34
Solution Stochiometry Problem:
48.0 mL of Ca(OH)2 solution was titrated with 19.2
mL of 0.385 M HNO3. Determine the molarity of
the Ca(OH)2 solution.
We must first
write a balanced
equation.
35
Solution Stochiometry Problem:
48.0 mL of Ca(OH)2 solution was titrated with 19.2
mL of 0.385 M HNO3. Determine the molarity of
the Ca(OH)2 solution.
Ca(OH)2(aq) + 2 HNO3(aq)  2 H2O(l) + Ca(NO3)2(aq)
48.0 mL
?M
HNO3
19.2 mL
19.2 mL
0.385 mol
0.385 M 
L
HNO 3
0.385 mol 1mol Ca(OH) 2
=0.0770 mol(Ca(OH)
1000 mL 2mol HNO 3 48.0 x 10-3L
L (Ca(OH) )
2)
2
HNO 3
units match!
36
37
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
b. Determine the limiting reactant.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
Nowthe
place
First copy down
numerical the
the BALANCED
information below
equation!
the compounds.
38
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
b. Determine the limiting reactant.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
0.15 mol
? moles
0.10 mol
Hide
Two starting
amounts?
Where do we
start?
one
39
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
b. Determine the limiting reactant.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
0.15 mol
? moles
Hide
0.10
mol
Based on:
0.15 mol KO2
KO2
3mol O 2
4mol KO 2
= 0.1125 mol O2
40
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
b. Determine the limiting reactant.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
0.15
mol
? moles
Hide
0.10 mol
Based on:
0.15 mol KO2
KO2
3mol O 2
4mol KO 2
Based on: 0.10 mol H2O 3mol O 2
H2 O
2mol H O
= 0.1125 mol O2
= 0.150 mol O2
2
41
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
Determine the limiting reactant.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
0.15 mol
? moles
0.10 mol
Based on:
0.15 mol KO2
KO2
Based on:
H2 O
3mol O 2
4mol KO 2
0.10 mol H2O 3mol O 2
H2O = excess (XS) reactant!
2mol H 2O
= 0.1125 mol O2
It was limited by the
amount of KO2.
= 0.150 mol O2
What is the theoretical yield?
Hint: Which is the smallest
amount? The is based upon the
limiting reactant?
42
Theoretical yield vs. Actual yield
Suppose the theoretical yield for an
experiment was calculated to be
19.5 grams, and the experiment was
performed, but only 12.3 grams of
product were recovered. Determine
the % yield.
Theoretical yield = 19.5 g based on limiting reactant
Actual yield = 12.3 g experimentally recovered
actual yield
% yield 
x 100
theoretica l yield
% yield 
12.3
x 100  63.1% yield
19.5
43
Limiting/Excess Reactant Problem with % Yield
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O,
how many grams of O2 can be produced?
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
?g
120.0 g
47.0 one
g
Hide
Based on: 120.0 g KO2
KO2
mol 3mol O 2 32.0g O 2
71.1g 4mol KO 2 mol O 2
= 40.51 g O2
44
Limiting/Excess Reactant Problem with % Yield
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O,
how many grams of O2 can be produced?
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
?g
120.0
47.0 g
Hideg
Based on: 120.0 g KO2
KO2
mol 3mol O 2 32.0g O 2
71.1g 4mol KO 2 mol O 2
mol H 2O
3 mol O 2 32.0g O 2
Based on: 47.0 g H2O
18.02 g H 2O 2 mol H 2O mol O 2
H2 O
= 40.51 g O2
= 125.3 g O2
Question if only 35.2 g of O2 were recovered, what was the percent yield?
actual
35.2
x 100 
x 100  86.9% yield
theoretica l
40.51
45
If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O,
how many grams of O2 can be produced?
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
?g
120.0 g
47.0 g
Based on: 120.0 g KO2
KO2
mol 3mol O 2 32.0g O 2
71.1g 4mol KO 2 mol O 2
= 40.51 g O2
mol H 2O
3 mol O 2 32.0g O 2
Based on: 47.0 g H2O
18.02 g H 2O 2 mol H 2O mol O 2
H2 O
= 125.3 g O2
Determine how many grams of Water were left over.
The Difference between the above amounts is directly RELATED to the XS H2O.
125.3 - 40.51 = 84.79 g of O2 that could have been formed from the XS water.
2 mol H 2O 18.02 g H 2O
32.0 g O 2 3 mol O 2 1 mol H 2O
84.79 g O2 mol O 2
= 31.83 g XS H2O
46
47
Try this problem (then check your answer):
Calculate the molarity of a solution prepared by dissolving 25.6 grams of
Al(NO3)3 in 455 mL of solution.
After you have
worked the
problem, click here
to see
setup answer
25.6 g mole
mol
 0.264
-3
213 g 455 x 10 L
L
48
49